+0

0
120
3

There are real numbers a and b such that for every positive number x, we have the identity

$$\tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right) + \tan^{-1}(ax) + \tan^{-1}(bx) = \frac{\pi}{2} \, .$$

What is the value of $$a^2 + b^2$$

Mar 10, 2020

#1
+3

Well, if I've interpreted the question correctly, then: Mar 10, 2020
#2
+3

Sorry Alan, that doesn't work. The identity has to be true for all x, not just x = 1/a.

Also, there's no reason why a should equal b.

Start by moving the first term to the rhs, and then take the tangent of both sides.

$$\displaystyle \tan(\tan^{-1}(ax)+\tan^{-1}(bx))=\tan\left(k-\tan^{-1}\left(\frac{1}{x}-\frac{x}{8}\right)\right)$$

(k for the moment, rather than pi/2).

Expand both sides using the tan(A + B) identity. Also, on the rhs, then divide top and bottom by tan(k).

$$\displaystyle \frac{ax+bx}{1-abx^{2}}=\frac{1-(1/x-x/8)/\tan(k)}{1/\tan(k)+(1/x-x/8)}.$$

Now replace k by pi/2 and this becomes

$$\displaystyle \frac{ax+bx}{1-abx^{2}}=\frac{1}{(1/x)-(x/8)}.$$

Cross multiply, multiply out and collect up terms,

$$\displaystyle a+b-\frac{x^{2}}{8}(a+b-8ab)=1.$$

For this to be an identity, true for all values of x, it's necessary that a + b - 8ab = 0, and also that a + b = 1.

$$\displaystyle a =\frac{2\pm \sqrt{2}}{4}\quad \text{and} \quad b=\frac{2 \mp \sqrt{2}}{4},$$

after which

$$\displaystyle a^{2}+b^{2}=\frac{3}{4}.$$

Mar 11, 2020
#3
0

Excellent!  So I did indeed misinterpret the question.

Alan  Mar 11, 2020